Question 2: Work, Energy and Power & Circular Motion

Question

A bead $P$ sliding on a frictionless semi-circular string $(A C B)$ and it is at point $S$ at $t =0$ and at this instant the horizontal component of its velocity is $v$. Another bead $Q$ of the same mass as $P$ is ejected from point $A$ at $t=0$ along the horizontal string $A B$, with the speed $v$, friction between the beads and the respective strings may be neglected in both cases. Let $t_P$ and $t_Q$ be the respective times taken by beads $P$ and $Q$ to reach the point $B$, then the relation between $t_P$ and $t_Q$ is

Accepted Answer

For bead Q
Bead $Q$ moves along the horizontal straight string $A B$. The length $A B$ is $2 R$ where $R$ is the radius of semicircular path.
Since the string is frictionless and horizontal, there is no component of gravity acting along the direction of motion.
Bead Q maintains a constant horizontal velocity v throughout the motion.
$$ t_Q=\frac{A B}{v}=2\left[\frac{R}{v}\right] $$
For bead P
Bead P moves along the semi-circular path ACB under the influence of gravity. And along horizontal the net force is zero, i.e., the acceleration along vertical will be non-zero and the acceleration along horizontal will be zero.
At point S , the horizontal component of its velocity is given as v .
As bead P moves from S toward the bottom-most point C , the vertical component of velocity changes but the horizontal component remains same.
When bead P reaches point B then the horizontal displacement is $\mathrm{S}^{\prime} \mathrm{O}+\mathrm{OB}$
$$ \mathrm{S}^{\prime} \mathrm{O}=\mathrm{R} \cos 45^{\circ} $$
Then the time taken to reach point $B$ is $t_P=\frac{S^{\prime} B}{v}=\frac{R \cos 45^{\circ}+R}{v}=\frac{R}{v}\left[\frac{1}{\sqrt{2}}+1\right]$
So, the ratio of time taken by $Q$ and $P$ to reach $B$ is,
$\frac{t_Q}{t_P}=\frac{2[R / v]}{\left[\frac{1}{\sqrt{2}}+1\right][R / v]}=\frac{2 \sqrt{2}}{\sqrt{2}+1}=[8-2 \sqrt{2}]>1$
$$\Rightarrow $$ $\frac{\mathrm{t}_{\mathrm{Q}}}{\mathrm{t}_{\mathrm{P}}}>1 \Rightarrow \mathrm{t}_{\mathrm{Q}}>\mathrm{t}_{\mathrm{P}}$

Answer

$t_P < t_Q$

Answer

$t_P=t_Q$

Answer

$t_P>1.25 t_Q$

Answer

$t_P>t_Q$

Work, Energy and Power & Circular Motion

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